NCERT Solutions · Class 12 Mathematics Inverse Trigonometric Functions

Let \(\displaystyle \theta=\sin^{-1}x\), so that \(\displaystyle x=\sin\theta\).Settle the range before the algebra. For \(\displaystyle x\in\left[-\frac{1}{2},\frac{1}{2}\right]\) we get \(\displaystyle \theta\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right]\), and therefore \[3\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\]By the triple-angle identity \(\displaystyle \sin 3\theta=3\sin\theta-4\sin^{3}\theta\), \[\sin 3\theta=3x-4x^{3}.\]The principal-value branch of \(\displaystyle \sin^{-1}\) is \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), and \(\displaystyle 3\theta\) has just been shown to lie in it, so \(\displaystyle \sin^{-1}\) undoes \(\displaystyle \sin\) here: \[\sin^{-1}\left(3x-4x^{3}\right)=3\theta=3\sin^{-1}x.\]Hence \(\displaystyle 3\sin^{-1}x=\sin^{-1}\left(3x-4x^{3}\right)\) for all \(\displaystyle x\in\left[-\frac{1}{2},\frac{1}{2}\right]\). The restriction is doing real work: once \(\displaystyle |x|>\frac{1}{2}\), \(\displaystyle 3\theta\) leaves the principal branch and the equality breaks.

Let \(\displaystyle \theta=\cos^{-1}x\), so that \(\displaystyle x=\cos\theta\).Range first. \(\displaystyle \cos^{-1}\) is decreasing, so \(\displaystyle x\in\left[\frac{1}{2},1\right]\) gives \(\displaystyle \theta\in\left[0,\frac{\pi}{3}\right]\), and therefore \[3\theta\in\left[0,\pi\right].\]By the triple-angle identity \(\displaystyle \cos 3\theta=4\cos^{3}\theta-3\cos\theta\), \[\cos 3\theta=4x^{3}-3x.\]The principal-value branch of \(\displaystyle \cos^{-1}\) is \(\displaystyle [0,\pi]\), and \(\displaystyle 3\theta\) lies in it, so \[\cos^{-1}\left(4x^{3}-3x\right)=3\theta=3\cos^{-1}x.\]Hence \(\displaystyle 3\cos^{-1}x=\cos^{-1}\left(4x^{3}-3x\right)\) for \(\displaystyle x\in\left[\frac{1}{2},1\right]\). Without the restriction \(\displaystyle 3\theta\) can exceed \(\displaystyle \pi\) and the identity fails.

Substitute \(\displaystyle x=\tan\theta\), i.e. \(\displaystyle \theta=\tan^{-1}x\), with \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) and \(\displaystyle \theta\neq 0\) (since \(\displaystyle x\neq 0\)).Then \(\displaystyle 1+x^{2}=\sec^{2}\theta\). On \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) we have \(\displaystyle \cos\theta>0\), so \(\displaystyle \sec\theta>0\) and the positive square root is \[\sqrt{1+x^{2}}=\sec\theta.\]Therefore, using \(\displaystyle 1-\cos\theta=2\sin^{2}\frac{\theta}{2}\) and \(\displaystyle \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\), \[\frac{\sqrt{1+x^{2}}-1}{x}=\frac{\sec\theta-1}{\tan\theta}=\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^{2}\dfrac{\theta}{2}}{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}=\tan\frac{\theta}{2}.\]Since \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), we have \(\displaystyle \frac{\theta}{2}\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\), which sits inside the principal branch of \(\displaystyle \tan^{-1}\), so \(\displaystyle \tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}\).\[\tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x}=\frac{1}{2}\tan^{-1}x,\qquad x\neq 0.\]

Use the half-angle forms \(\displaystyle 1-\cos x=2\sin^{2}\frac{x}{2}\) and \(\displaystyle 1+\cos x=2\cos^{2}\frac{x}{2}\): \[\frac{1-\cos x}{1+\cos x}=\frac{2\sin^{2}\dfrac{x}{2}}{2\cos^{2}\dfrac{x}{2}}=\tan^{2}\frac{x}{2}.\]Now take the square root, and this is the step to be careful with: \(\displaystyle \sqrt{t^{2}}=|t|\). Given \(\displaystyle 0<x<\pi\), we have \(\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{2}\right)\), where \(\displaystyle \tan\frac{x}{2}>0\), so \[\sqrt{\tan^{2}\frac{x}{2}}=\tan\frac{x}{2}.\]Since \(\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{2}\right)\subset\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), the principal branch of \(\displaystyle \tan^{-1}\) returns the angle itself: \[\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2},\qquad 0<x<\pi.\]

Do not divide by \(\displaystyle \cos x\) here: \(\displaystyle x=\frac{\pi}{2}\) lies in the given interval. Write both parts as single trigonometric functions instead: \[\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right),\qquad \cos x+\sin x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right).\]For \(\displaystyle -\frac{\pi}{4}<x<\frac{3\pi}{4}\) we have \(\displaystyle x+\frac{\pi}{4}\in(0,\pi)\), so \(\displaystyle \sin\left(x+\frac{\pi}{4}\right)>0\) and the quotient is defined throughout. Hence \[\frac{\cos x-\sin x}{\cos x+\sin x}=\cot\left(x+\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{2}-x-\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{4}-x\right).\]Check the branch: \(\displaystyle -\frac{\pi}{4}<x<\frac{3\pi}{4}\) gives \(\displaystyle \frac{\pi}{4}-x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), the principal branch of \(\displaystyle \tan^{-1}\). Therefore \[\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x,\qquad -\frac{\pi}{4}<x<\frac{3\pi}{4}.\]

Since \(\displaystyle |x|<a\), necessarily \(\displaystyle a>0\). Substitute \(\displaystyle x=a\sin\theta\), i.e. \(\displaystyle \theta=\sin^{-1}\frac{x}{a}\); as \(\displaystyle \left|\frac{x}{a}\right|<1\), \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).Then \(\displaystyle a^{2}-x^{2}=a^{2}\cos^{2}\theta\), and since \(\displaystyle \cos\theta>0\) on that interval the positive root is \(\displaystyle \sqrt{a^{2}-x^{2}}=a\cos\theta\). So \[\frac{x}{\sqrt{a^{2}-x^{2}}}=\frac{a\sin\theta}{a\cos\theta}=\tan\theta.\]Because \(\displaystyle \theta\) already lies in \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[\tan^{-1}\frac{x}{\sqrt{a^{2}-x^{2}}}=\theta=\sin^{-1}\frac{x}{a},\qquad |x|<a.\]

Divide numerator and denominator by \(\displaystyle a^{3}\) (legitimate, \(\displaystyle a>0\)): \[\frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}=\frac{3\left(\dfrac{x}{a}\right)-\left(\dfrac{x}{a}\right)^{3}}{1-3\left(\dfrac{x}{a}\right)^{2}}.\]Put \(\displaystyle \frac{x}{a}=\tan\theta\), i.e. \(\displaystyle \theta=\tan^{-1}\frac{x}{a}\). By the triple-angle formula for the tangent, \[\tan 3\theta=\frac{3\tan\theta-\tan^{3}\theta}{1-3\tan^{2}\theta},\] so the expression inside \(\displaystyle \tan^{-1}\) is exactly \(\displaystyle \tan 3\theta\).Range check, which is what the given interval is for: \(\displaystyle -\frac{a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}\) means \(\displaystyle \left|\tan\theta\right|<\frac{1}{\sqrt{3}}\), so \(\displaystyle |\theta|<\frac{\pi}{6}\) and hence \(\displaystyle 3\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). (The same interval keeps \(\displaystyle 1-3\tan^{2}\theta\neq 0\), i.e. the denominator \(\displaystyle a\left(a^{2}-3x^{2}\right)\neq 0\).)Therefore \[\tan^{-1}\left(\frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}\right)=3\theta=3\tan^{-1}\frac{x}{a}.\]

Work from the inside out, using principal values at each stage.\(\displaystyle \sin^{-1}\frac{1}{2}\) is the angle in \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) whose sine is \(\displaystyle \frac{1}{2}\), so \(\displaystyle \sin^{-1}\frac{1}{2}=\frac{\pi}{6}\). Then \[2\sin^{-1}\frac{1}{2}=\frac{\pi}{3},\qquad \cos\frac{\pi}{3}=\frac{1}{2},\qquad 2\cos\left(2\sin^{-1}\frac{1}{2}\right)=2\cdot\frac{1}{2}=1.\]Finally \(\displaystyle \tan^{-1}1\) is the angle in \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) with tangent \(\displaystyle 1\): \[\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right]=\tan^{-1}1=\frac{\pi}{4}.\]

Convert both inverse functions to \(\displaystyle \tan^{-1}\) using the standard identities \[\sin^{-1}\frac{2x}{1+x^{2}}=2\tan^{-1}x\ \ (|x|\le 1),\qquad \cos^{-1}\frac{1-y^{2}}{1+y^{2}}=2\tan^{-1}y\ \ (y\ge 0).\] The given conditions \(\displaystyle |x|<1\) and \(\displaystyle y>0\) are exactly what makes each identity applicable (for \(\displaystyle x>1\) or \(\displaystyle y<0\) the right-hand sides would need a correction term).Hence the bracket equals \(\displaystyle 2\tan^{-1}x+2\tan^{-1}y\), and half of it is \[\frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^{2}}+\cos^{-1}\frac{1-y^{2}}{1+y^{2}}\right]=\tan^{-1}x+\tan^{-1}y.\]Write \(\displaystyle A=\tan^{-1}x\), \(\displaystyle B=\tan^{-1}y\), so \(\displaystyle \tan A=x\), \(\displaystyle \tan B=y\), and apply the addition formula \[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{x+y}{1-xy},\] which is finite because \(\displaystyle xy<1\).\[\tan\frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^{2}}+\cos^{-1}\frac{1-y^{2}}{1+y^{2}}\right]=\frac{x+y}{1-xy}.\]

\(\displaystyle \sin^{-1}(\sin\theta)=\theta\) only when \(\displaystyle \theta\) lies in the principal branch \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Here \(\displaystyle \frac{2\pi}{3}>\frac{\pi}{2}\), so the answer is not \(\displaystyle \frac{2\pi}{3}\); replace the angle by one in the branch with the same sine.Using \(\displaystyle \sin(\pi-\theta)=\sin\theta\), \[\sin\frac{2\pi}{3}=\sin\left(\pi-\frac{2\pi}{3}\right)=\sin\frac{\pi}{3},\qquad \frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\]Therefore \[\sin^{-1}\left(\sin\frac{2\pi}{3}\right)=\frac{\pi}{3}.\]